The number of 4-letter words, with or without meaning, which can be formed using the letters PQRPQRSTUVP, is ___ .

The word given is PQRPQRSTUVP. Here, we have P 3 times, Q 2 times, R 3 times, and S,T,U,V one time each
Total distinct letters: 7

We form 4-letter words using these.

Case 1: All letters are distinct

Here, 4 distinct letters can be chosen from 7 letters in  $$ {}^7C_4 = 7 \cdot 6 \cdot 5 \cdot 4 = 840 \text { ways} $$

Case 2: One pair + two distinct

To choose a pair, we need to pick one from the letters that occur 2 times or more in the word. Of the letters P,Q and R occure more than once, so the letter that will form the pair can be picked in 3 ways. For the remaining 2 distinct letters, they can be picked from the remaining 6 letters in  $$  \binom{6}{2} = 15 $$ ways. These can be arranged amongst themselves in  $$  \frac{4!}{2!} = 12$$ ways

Total number of ways words can be formed is= $$ 3 \cdot 15 \cdot 12 = 540 \text { ways} $$

Case 3: Two pairs

Here, we need to choose 2 letters (each must occur ≥2 times): from (P,Q,R), which can be done in 3 ways. The letters can be arranged amongst themselves in $$ \dfrac{4!}{2!2!} = 6 $$ ways

Total number of ways words can be formed is $$ 3 \cdot 6 = 18 \text{ ways} $$

Case 4: Three same + one different

Only P occurs three times, so here it can be picked in 1 way, and the 1 different letter from the remaining 6 can be picked in 6 ways. The letters can be arranged amongst themselves in  $$ \dfrac{4!}{3!} = 4  $$

Total number of ways words can be formed is $$ 1 \cdot 6 \cdot 4 = 24 \text{ ways}$$

Hence, the total number of ways words can be formed considering all the cases is  $$ 840 + 540 + 18 + 24 =1422 \text { ways} $$